Question

Two bodies of mass \( 1 \) kg and \( 3 \) kg have position vectors \( \hat{i} + 2\hat{j} + \hat{k} \) and \( -3\hat{i} - 2\hat{j} + \hat{k} \) respectively. The magnitude of the position vector of the center of mass of this system will be similar to the magnitude of which vector?

• A. \( \hat{i} - 2\hat{j} + \hat{k} \)
• B. \( -3\hat{i} - 2\hat{j} + \hat{k} \)
• C. \( -2\hat{i} + 2\hat{k} \)
• D. \( -2\hat{i} - \hat{j} + 2\hat{k} \)

Hint:

The center of mass position is given by \( \vec{r}_{{com}} = \frac{\sum m_i \vec{r}_i}{\sum m_i} \). To find the correct vector, compare magnitudes.

Answer 1

The Correct Option is A

Step 1: {Formula for center of mass} 
\[ \vec{r}_{{com}} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \] 
Step 2: {Substituting values} 
\[ \vec{r}_{{com}} = \frac{(1)(\hat{i} + 2\hat{j} + \hat{k}) + (3)(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3} \] \[ = \frac{\hat{i} + 2\hat{j} + \hat{k} - 9\hat{i} - 6\hat{j} + 3\hat{k}}{4} \] \[ = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4} \] \[ = -2\hat{i} - \hat{j} + \hat{k} \] 
Step 3: {Find the magnitude} 
\[ |\vec{r}_{{com}}| = \sqrt{(-2)^2 + (-1)^2 + (1)^2} \] \[ = \sqrt{4 + 1 + 1} = \sqrt{6} \] The only vector with the same magnitude is \( \hat{i} - 2\hat{j} + \hat{k} \). 
 

Answer 2

Step 1: Let the position vector of mass \( m_1 = 1\,kg \) be:
\( \vec{r}_1 = \hat{i} + 2\hat{j} + \hat{k} \)
Let the position vector of mass \( m_2 = 3\,kg \) be:
\( \vec{r}_2 = -3\hat{i} - 2\hat{j} + \hat{k} \)

Step 2: Use the center of mass formula:
\( \vec{R}_{cm} = \frac{m_1 \vec{r}_1 + m_2 \vec{r}_2}{m_1 + m_2} \)
Substitute the values:
\( \vec{R}_{cm} = \frac{1(\hat{i} + 2\hat{j} + \hat{k}) + 3(-3\hat{i} - 2\hat{j} + \hat{k})}{1 + 3} \)

Step 3: Simplify the numerator:
\( = \hat{i} + 2\hat{j} + \hat{k} - 9\hat{i} - 6\hat{j} + 3\hat{k} = (-8\hat{i} - 4\hat{j} + 4\hat{k}) \)

Step 4: Divide by 4:
\( \vec{R}_{cm} = \frac{-8\hat{i} - 4\hat{j} + 4\hat{k}}{4} = -2\hat{i} - \hat{j} + \hat{k} \)

Step 5: Find the magnitude of this vector:
\( |\vec{R}_{cm}| = \sqrt{(-2)^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6} \)

Step 6: Now find the magnitude of the given options.
Check for vector \( \hat{i} - 2\hat{j} + \hat{k} \):
\( |\vec{A}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \)

Final Answer: The magnitude of the position vector of the center of mass is same as the magnitude of \( \hat{i} - 2\hat{j} + \hat{k} \)